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kodGreya [7K]
3 years ago
7

A jet pilot takes his aircraft into a vertical loop. If the jet is moving at a speed of 700km/hr at the lowest point in the loop

, determine the radius of the circle so that the acceleration which the pilot experiences at the lowest point of the loop does not exceed 6 time the acceleration due to gravity or (6g’s).
Physics
1 answer:
RideAnS [48]3 years ago
7 0

Answer:

<u>The radius of circle is 643 m. </u>

Explanation:

The centripetal acceleration of the jet is

a=\frac{V^{2}}{r}

Where,

a is "centripetal acceleration"

V is "speed "

r is "radius of the circle "

Given values ,

Speed of the jet (v) = 700 km/h

(1 km = 1000 m; 1 h = 3600 sec. To convert km/h into m/sec, multiply the number by 5 and then divide it by 18.)

Speed of the jet (v)=\left(700 \times \frac{5}{18}\right) \mathrm{m} / \mathrm{s}

V = 194.444 m/s

a = 6g's

\text { (g is referred to as the acceleration of gravity. Its value is }\left.9.8 \mathrm{m} / \mathrm{s}^{2} \text { on Earth }\right)

a=6 \times 9.8 \mathrm{m} / \mathrm{s}^{2}

a=58.8 \mathrm{m} / \mathrm{s}^{2}

\text { The radius of circle is a }=\frac{V^{2}}{r}

Substitute the given values in the formula,

a=\frac{194.444^{2}}{58.8}

a=\frac{37808.46}{58.8}

a = 643 m

The radius of circle is 643 m.

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